Codeforces Round #436 (Div. 2) B.Polycarp and Letters

因为难得又一次CF的比赛是非常清真的傍晚，超级少见啊

所以当然要打啦，于是rank:87，rating+=76，滞留在上紫的边缘

下面把几道觉得还不错的题目来总结一下

B.Polycarp and Letters

Polycarp loves lowercase letters and dislikes uppercase ones. Once he got a string s consisting only of lowercase and uppercase Latin letters.

Let A be a set of positions in the string. Let's call it pretty if following conditions are met:

• letters on positions from A in the string are all distinct and lowercase;
• there are no uppercase letters in the string which are situated between positions from A (i.e. there is no such j that s[j] is an uppercase letter, and a1 < j < a2 for some a1 and a2 from A).

Write a program that will determine the maximum number of elements in a pretty set of positions.

input

11
aaaaBaabAbA

output

2

input

12
zACaAbbaazzC

output

3

Note

In the first example the desired positions might be 6 and 8 or 7 and 8. Positions 6 and 7 contain letters 'a', position 8 contains letter 'b'. The pair of positions 1 and 8 is not suitable because there is an uppercase letter 'B' between these position.

In the second example desired positions can be 7, 8 and 11. There are other ways to choose pretty set consisting of three elements.

一句话题意：让你选择最多的字母，使得这些字母都是不同的小写字母，并且它们在原串的位置之间没有大写字母。

我觉得就以这道题的难度都可以做A了吧

总之就是暴力，反正数据范围小，就算n^3的都没事（我打的就是这种辣鸡算法）

先枚举头和尾，然后在用bool数组来记录出现了那些小写字母，

然后的然后就可以A了啊，完全没有坑.

#include<bits/stdc++.h>
using namespace std;
int n,ans,res;
char s[1000];
bool bo[1000];
int main(){
    scanf("%d%s",&n,s+1);
    for (int i=1;i<=n;++i)
        for (int j=1;j<=n;++j){
            ans=0; memset(bo,0,sizeof(bo));
            for (int k=i;k<=j;++k){
                if (s[k]>='A'&&s[k]<='Z'){
                    ans=0; break;
                }
                if (!bo[s[k]]) ans++,bo[s[k]]=1;
                res=max(res,ans);
            }
        }
    printf("%d",res);    
}